If a circuit has a 9V battery and two resistors of 3Ω and 6Ω in series, what is

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Question: If a circuit has a 9V battery and two resistors of 3Ω and 6Ω in series, what is the voltage across the 6Ω resistor?

Options:

Correct Answer: 6V

Solution:

Using the voltage divider rule, V6Ω = (R6Ω / (R3Ω + R6Ω)) * Vtotal = (6 / (3 + 6)) * 9 = 6V.

If a circuit has a 9V battery and two resistors of 3Ω and 6Ω in series, what is the voltage across the 6Ω resistor?

If a circuit has a 9V battery and two resistors of 3Ω and 6Ω in series, what is the voltage across the 6Ω resistor?

Correct Answer: 6V

Step 1: Identify the total voltage from the battery, which is 9V.
Step 2: Identify the resistors in the circuit. We have a 3Ω resistor and a 6Ω resistor in series.
Step 3: Calculate the total resistance in the circuit by adding the two resistors together: 3Ω + 6Ω = 9Ω.
Step 4: Use the voltage divider rule to find the voltage across the 6Ω resistor. The formula is: V6Ω = (R6Ω / (R3Ω + R6Ω)) * Vtotal.
Step 5: Substitute the values into the formula: V6Ω = (6 / (3 + 6)) * 9.
Step 6: Simplify the equation: V6Ω = (6 / 9) * 9.
Step 7: Calculate the result: V6Ω = 6V.

Voltage Divider Rule – The voltage divider rule is used to determine the voltage drop across a resistor in a series circuit based on the resistance values and the total voltage.
Series Circuit – In a series circuit, the total resistance is the sum of individual resistances, and the same current flows through all components.